Circle to circle homotopic to the constant map?

For the first part, let $i\colon S^1\to D^2$ be the inclusion of the circle into the unit disk and, since $f$ is null-homotopic, let $\tilde{f}\colon D^2\to S^1$ be an extension of $f$ to the whole disk (which exists). Since $f$ has no fixed points, and the image of $\tilde{f}$ lies within $S^1$, what can we say about $i\circ \tilde{f}\colon D^2\to D^2$ and what theorem about maps on disks does this contradict?

For the second part, just prove that the composition of a nullhomotopic map with the map which rotates the circle by $\pi$ is also nullhomotopic (hint: rotation is homotopic to the identity and if $f\simeq f'$ and $g\simeq g'$ then $f\circ g\simeq f'\circ g'$), and then use part a.

Lemma: Show that if $A$ is a retract of $B^2$, then every continuous map $f : A \to A$ has a fixed point.

Proof: Suppose that $A$ is a retract of $B^2$, then by definition there exists a continuous map $r : B^2 \to A$ such that $r(a) = a$ for all $a \in A$. Let $f : A \to A$ be an arbitrary continuous map. Define $g : B^2 \to B^2$ by $g = j \circ f \circ r$ where $j : A \to B^2$ is the inclusion map. By the Brouwer fixed-point theorem for the disk, there exists $x \in B^2$ such that $g(x) = x$. But notice that $g(x) = j(f(r(x))) = f(r(x)) = x$ which means that $x \in A$ since $x \in \operatorname{Im}(f) \subseteq A$. Since $r$ is a retraction of $B^2$ onto $A$, $r(x) = x$ and hence $f(r(x)) = f(x) = x$. Conclude that $f$ has a fixed point.

Theorem: Show that if $h : S^1 \to S^1$ is nulhomotopic, then $h$ has a fixed point and $h$ maps some point $x$ to its antipode $-x$.

Proof: Since $h : S^1 \to S^1$ is nulhomotopic there exists a continuous extension $k : B^2 \to S^1$ of $h$ into $B^2$. Define $g : B^2 \to B^2$ by $g = j \circ k$ where $j : S^1 \to B^2$ is the inclusion map. $g$ is continuous, so by the fixed point theorem, there exists a fixed point $x \in B^2$ such that $g(x) = x$. But notice that $x = g(x) = j(k(x)) = k(x) \in S^1$ so $x \in S^1$ and hence $k(x) = h(x) = x$ and thus $h$ has a fixed point.

Define $\alpha : S^1 \to S^1$ by $\alpha(x) = -x$. By hypothesis, $h$ is nulhomotopic, so there exists $c \in S^1$ such that $h$ is homotopic to $e_c$. In particular, there exists a homotopy $F : S^1 \times I \to S^1$ such that $F(s, 0) = h(s)$ and $F(s, 1) = e_c(s) = c$. Since $\alpha$ is continuous, then $\alpha \circ F$ is a homotopy between $\alpha \circ h$ and $\alpha \circ e_c = e_{-c}$. Hence $\alpha \circ h$ is nulhomotopic. By previous discussion, there exists a fixed point $x$ such that $\alpha(h(x)) = -h(x) = x$. Multiply both sides by -1 and we get $h(x) = -x$. Conclude that $h$ maps some point $x$ to its antipode $-x$.